这道题只要把双向链表和哈希表连接起来就行,只是写起来有点麻烦了
java
class LRUCache {
class DLinkedNode {
int key;
int value;
DLinkedNode prev;
DLinkedNode next;
public DLinkedNode(){}
public DLinkedNode(int _key, int _value) {key = _key;value = _value;}
}
private Map<Integer, DLinkedNode> cache = new HashMap<Integer, DLinkedNode>();
private int size;
private int capacity;
private DLinkedNode head, tail;
public LRUCache(int capacity) {
this.size = 0;
this.capacity = capacity;
//使用伪头部和伪尾部
head = new DLinkedNode();
tail = new DLinkedNode();
//双向链表
head.next = tail;
tail.prev = head;
}
public int get(int key) {
DLinkedNode node = cache.get(key);
if(node == null){
return -1;
}
//如果节点存在
moveToHead(node);
return node.value;
}
public void put(int key, int value) {
//尝试从原来的双向哈希链表中获取
DLinkedNode node = cache.get(key);
if(node == null){
//不存在创建新节点
DLinkedNode newNode = new DLinkedNode(key, value);
//添加进表
cache.put(key,newNode);
//添加到双向链表头部
addToHead(newNode);
++size;
if (size > capacity){
//超出容量,删除尾部元素
DLinkedNode tail = removeTail();
//删除哈希表中对应项
cache.remove(tail.key);
--size;
}
}
//存在节点,先定位再修改数据,并移到头部
else{
node.value = value;
moveToHead(node);
}
}
private void moveToHead(DLinkedNode node){
removeNode(node);
addToHead(node);
}
private void removeNode(DLinkedNode node){
node.prev.next = node.next;
node.next.prev = node.prev;
}
private void addToHead(DLinkedNode node){
node.prev = head;
node.next = head.next;f
head.next.prev = node;
head.next = node;
}
private DLinkedNode removeTail(){
DLinkedNode res = tail.prev;
removeNode(res);
return res;
}
}
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/